Recent content by MichaelWiteman

Hi, To compute, for example, the Gibbs energy change for a ligand binding to a protein, various so called alchemical methods are used in molecular dynamics simulations. My question is why can't we just obtain averaged Gibbs energies for 1) the free ligand and protein in the same water box, and...

I guess I skipped a step for my own convenience since I didn't know how to formally show it. I removed the 2nd term in the rhs of your equation for Wg(t). The v in that term will be changing direction (oscillation) and magnitude (damping) with time until v=0. The way I see it now, is that work...

So Wg(t) = -mg dx, If mg = F = PA, then W=-PA dx = - P dv. Since P is here the external pressure (its derived from mg) we arrived at the equation from my textbook: Wirrev = -Pext dV Cool!

In a case with an ideal gas I believe there will be no damping and the piston with the weight will oscillate forever. However, in a real situation the gas will have some viscosity and the damping mechanism would be via internal friction.

I'm not a physicist so I apologize if I write something that you might find ignorant. The gas has pressure and the piston has area, hence there is a force acting on it equal to F(t)=P(t)*A. The piston with the weight exerts a force on the gas equal to mg. So the net force is equal to F(t)-mg...

Sorry, I meant the total energy change is zero. --Mike

Thank you for your reply! I should had mentioned that this is in vacuum, but it doesn't invalidate your answer. If I understand correctly, you are using the fact the total total energy (where the system is the expanding gas, and surroundings is the moving weight) is equal to zero and that the...

Hi, An irreversible gas expansion is often described in textbooks with a compressed gas in a cylinder pushing up a weight (with mass m) via a hypothetical friction-less and weightless piston. It is said the work done by the gas is equal to -mg × h and from this you can derive the work for a...

Chestermiller, thank you for your reply! I understand now were my logic broke. I was fixated on the the gas that is expanding in the left chamber and ignored the gas in the right chamber, which is being compressed as the system equilibrates, and thus the overall work in the system is zero. This...

Hi, Joule's original gas expansion experiment is often presented like in the following link: http://demonstrations.wolfram.com/JouleExperimentOnFreeExpansion/ The apparent lack of temperature change in this experiment is often used in textbooks to demonstrate that the energy of an ideal gas is...